Question Number 78460 by Tony Lin last updated on 17/Jan/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }=? \\ $$ Commented by mathmax by abdo last updated on…
Question Number 78449 by mathocean1 last updated on 17/Jan/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{definition}\:\mathrm{of}\: \\ $$$$\mathrm{f}\left({x}\right)=\frac{−{x}}{\mid{x}\mid−{x}} \\ $$ Commented by Rio Michael last updated on 18/Jan/20 $${f}\left({x}\right)\:=\:\frac{{x}}{{x}−\mid{x}\mid} \\ $$$${so}\:{x}\:<\:\mathrm{0}\:{so}\:{that}\:{x}−\mid{x}\mid\:{should}\:{be}\:…
Question Number 143980 by mathdanisur last updated on 20/Jun/21 Answered by mitica last updated on 20/Jun/21 $${p}={a}+{b}+{c};{q}={ab}+{bc}+{ac};{r}={abc}\Rightarrow{q}^{\mathrm{2}} \geqslant\mathrm{3}{pr} \\ $$$$\mathrm{3}\Sigma\frac{{a}}{{b}}−\Sigma\frac{\mathrm{3}{a}+{b}+{c}}{{b}+{c}}=\mathrm{3}\Sigma\left(\frac{{a}}{{b}}−\frac{{a}}{{b}+{c}}\right)−\mathrm{3}= \\ $$$$\mathrm{3}\Sigma\frac{{ac}}{{b}\left({b}+{c}\right)}−\mathrm{3}=\frac{\mathrm{3}}{{abc}}\Sigma\frac{\left({ac}\right)^{\mathrm{2}} }{{b}+{c}}−\mathrm{3}\geqslant \\ $$$$\frac{\mathrm{3}}{{r}}\centerdot\frac{{q}^{\mathrm{2}}…
Question Number 12909 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 Answered by nume1114 last updated on 07/May/17 $$\:\:\:\:{a}^{\mathrm{2}} \left({a}+\mathrm{1}\right)+{b}^{\mathrm{2}} \left({b}+\mathrm{1}\right)+\mathrm{5}{ab} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{5}{ab}…
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Question Number 143979 by mathdanisur last updated on 20/Jun/21 Answered by mitica last updated on 20/Jun/21 $${p}={x}+{y}+{z};{q}={xy}+{yz}+{xz};{r}={xyz}\Rightarrow{p}^{\mathrm{2}} \geqslant\mathrm{3}{q} \\ $$$$\mathrm{3}\left({p}+\frac{\mathrm{3}{r}}{{q}}\right)^{\mathrm{4}} =\mathrm{3}\left(\frac{{p}}{\mathrm{3}}+\frac{{p}}{\mathrm{3}}+\frac{{p}}{\mathrm{3}}+\frac{\mathrm{3}{r}}{{q}}\right)^{\mathrm{4}} \overset{{am}−{gm}} {\geqslant} \\ $$$$\geqslant\mathrm{3}\left(\mathrm{4}\centerdot\sqrt[{\mathrm{4}}]{\frac{{p}}{\mathrm{3}}\centerdot\frac{{p}}{\mathrm{3}}\centerdot\frac{{p}}{\mathrm{3}}\centerdot\frac{\mathrm{3}{r}}{{q}}}\right)^{\mathrm{4}}…
Question Number 78440 by ajfour last updated on 17/Jan/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}\:=\:{p} \\ $$$$\:\:\left({a}−\mathrm{2}{b}\right)\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)=−\mathrm{24}{q} \\ $$$${Find}\:\:\:\frac{{a}+{b}}{\mathrm{3}}\:\:\:\:{in}\:{terms}\:{of}\:{p},\:{q}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12889 by sin (x) last updated on 05/May/17 $${f}\left({x}−\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{3}}+{f}\left({x}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{36}−{f}\left(\mathrm{21}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)=? \\ $$ Commented by prakash jain last updated on 06/May/17…
Question Number 12885 by FilupS last updated on 05/May/17 $${f}\left({f}\left({x}\right)\right)={f}^{\mathrm{2}} \left({x}\right) \\ $$$$\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}? \\ $$ Answered by mrW1 last updated on 05/May/17 $${u}={f}\left({x}\right)…
Question Number 12883 by FilupS last updated on 05/May/17 $$\mathrm{for}\:\:\:\:\:−\mathrm{128}\leqslant{x}\leqslant\mathrm{127} \\ $$$$\mathrm{and}\:\:\:−\mathrm{127}\leqslant{y}\leqslant\mathrm{128} \\ $$$$\mathrm{where}\:\:\:{x},{y}\in\mathbb{Z} \\ $$$$\: \\ $$$$\mathrm{Point}\:{P}\left({x},{y}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{cartesian}\:\mathrm{plane}. \\ $$$$\: \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{angle}\:\theta\:\mathrm{is}\:\mathrm{made}\:\mathrm{counter} \\…